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深度学习文章阅读（广义TS模糊系统）

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作者 sirlis 29 分钟阅读

本文介绍了 2001 年 Taniguchi 等提出的一种广义 TS 模糊系统的建模方法、规则约减和鲁棒控制方法。

T. Taniguchi; K. Tanaka; H. Ohtake; H.O. Wang. Model construction, rule reduction, and robust compensation for generalized form of Takagi-Sugeno fuzzy systems. IEEE Transactions on Fuzzy Systems ( Volume: 9, Issue: 4, Aug 2001).

1. 引言

在线性矩阵不等式（linear matrix inequality, LMI）设计框架下，基于 TS 模糊模型的非线性控制得以广泛应用。一般分为三个阶段：

  • 第一阶段：对非线性被控对象的模糊建模
    • 利用输入输出数据进行模糊模型辨识（Takagi and Sugeno, 1993 等）
    • 或 基于分区非线性思想的模糊系统构建（模糊 IF-THEN 规则）
  • 第二阶段：模糊控制规则推导，它反映了模糊模型的规则结构，它通过所谓的并行分布式补偿（PDC）实现
  • 第三阶段：模糊控制器设计，即确定反馈增益。

This paper presents a systematic procedure of fuzzy control system design that consists of fuzzy model construction, rule reduction, and robust compensation for nonlinear systems.

本文提出了一种模糊控制系统设计的系统程序，该程序由模糊模型构建，规则约简和非线性系统的鲁棒补偿组成。

注意，在本篇文章之前，还有两篇关键文章作为前续研究基础：

  • H. O. Wang, K. Tanaka, and M. Griffin, “Parallel distributed compensation of nonlinear systems by Takagi-Sugeno fuzzy model,” in Proceedings of 1995 IEEE International Conference on Fuzzy Systems. The International Joint Conference of the Fourth IEEE International Conference on Fuzzy Systems and The Second International Fuzzy Engineering Symposium, Yokohama, Japan, 1995, vol. 2, pp. 531–538, doi: 10.1109/FUZZY.1995.409737. 首次将TS模糊系统用于非线性系统的近似，给出了模糊系统 Lyapunov 稳定性的充分条件，研究了并行分布式补偿（PDC），给出了状态反馈控制律下的稳定性判别准则。
  • D. Localmodel, “Stability Analysis of Fuzzy Control Systems,” IEEE TRANSACTIONS ON SYSTEMS, MAN, AND CYBERNETICS-PART B: CYBERNETICS, vol. 26, no. 1, p. 4, 1996. 提出了模糊状态反馈控制器，貌似放宽了稳定性的条件（还没细看）。

2. 传统 TS 模糊系统

形式如下

\[\begin{aligned} &rule\ i\quad (i=1,2,\cdots,r):\\ &{\rm IF}\ z_1(t)\ is\ M_{1i}\ and\ \cdots\ and\ z_p(t)\ is\ M_{pi}\\ &{\rm THEN}\ \dot \boldsymbol x(t) = \boldsymbol A_i \boldsymbol x(t) + \boldsymbol B_i \boldsymbol u(t) \end{aligned}\]

其中

  • $r$ 是规则个数
  • $M_{ji}$ 是第 $j$ 个输入分量的模糊集
  • $\boldsymbol x\in \mathbb R^n$ 是状态量，$\boldsymbol u\in \mathbb R^m$ 是输入量
  • $\boldsymbol A_i\in \mathbb R^{n\times n}, \boldsymbol B_i\in \mathbb R^{n\times m}$ 是系数矩阵
  • $z_1(t),\cdots,z_p(t)$ 是已知的前提变量，可能为可测量的状态量的函数、外部干扰，和/或时间，用 $\boldsymbol z(t)$ 来表示所有

给定一对 $[\boldsymbol x(t),\boldsymbol u(t),\boldsymbol z(t)]$，采用重心法（加权平均法）可以得到模糊系统的最终输出：

\[\dot \boldsymbol x(t) = \sum_{i=1}^r h_i(\boldsymbol z(t))(\boldsymbol A_i \boldsymbol x(t) + \boldsymbol B_i \boldsymbol u(t))\]

其中

\[\begin{aligned} h_i(\boldsymbol z(t)) &= \frac{\omega_i(\boldsymbol z(t))}{\sum_{i=1}^r\omega_i(\boldsymbol z(t))}\\ \omega_i(z(t)) &= \prod_{j=1}^p M_{ji}(z_j(t)) \end{aligned}\]

$h_i(\boldsymbol z(t))$ 是每条规则的归一化权重，$M_{ji}(z_j(t)$ 是第 $i$ 条规则中第 $j$ 个分量的模糊集 $M_{ji}$ 的隶属度值。

参考：单点模糊产生器、乘积推理机和中心平均解模糊器。

Wang H., Tanaka K. and Griifn M., Parallel distributed conpensation of nonlinear systems by Takagi and Sugeno’s fuzzy model, 1995, in Porc. 4 th IEEE Int. Conf. Fuzzy syst., Yokohama, Japan, pp: 531-538.

定理：当模糊规则条数适当时，模糊系统可以以任意的精度逼近实际的任意线性或非线性系统。（万能逼近器？）

MamdaniE.H.and Assilian 5., Applications of fuzzy algorithms for control of simple dynamic Plant, IEEE Proc. Part-D, 1974, vol. 121, no. 8, pp: 1585-1588

3. 广义 TS 模糊系统

个人前言：传统的 TS 模糊系统是有规则个数的概念的，但是下文作者提出的广义 TS 模糊系统不再强调规则的概念，而是直接对状态方程进行模糊近似。

3.1. 建模

3.1.1. 系统建模

考虑某一类非线性系统表示如下：

\[\dot x_i(t) = \sum_{j=1}^n f_{ij}(\boldsymbol z(t))x_j(t) + \sum_{k=1}^m g_{ik}(\boldsymbol z(t))u_k(t)\]

其中：

  • $x_1(t)\cdots x_n(t)$ 是状态量，$u_1(t)\cdots u_m(t)$ 是输入量
  • $z_1(t),\cdots,z_n(t)$ 是已知的变量，可能为状态量的函数、外部变量，和/或时间
  • $f_{ij}(\boldsymbol z(t)), g_{ik}(\boldsymbol z(t))$ 是关于 $\boldsymbol z(t)$ 矩阵

注意，上式中的 $j$ 是遍历所有状态量，$j=1,2,\cdots,n$，$k$ 是遍历所有输入量，$k=1,2,\cdots,m$。上式可以看作描述系统的非线性状态方程。该方程刻画了每个状态量的一阶导与所有状态量和控制（输入）量的线性组合关系。非线性可以体现在：$z_i(t)$ 可为 $x_i(t)$ 的非线性函数。假设 $z_i(t) = sin(x_i(t))$ ，$f_{ij}(\boldsymbol z(t))=\boldsymbol z(t) = sin(\boldsymbol x(t))$，则原始状态方程是关于 $\boldsymbol x(t)$ 的非线性方程组。

定义如下的新变量（表示系数 $f_{ij},g_{ik}$ 的最大最小值）

\[\begin{aligned} a_{ij1} &\equiv \mathop{\rm max}\limits_{\boldsymbol z(t)} f_{ij}(\boldsymbol z(t))\\ a_{ij2} &\equiv \mathop{\rm min}\limits_{\boldsymbol z(t)} f_{ij}(\boldsymbol z(t))\\ b_{ik1} &\equiv \mathop{\rm max}\limits_{\boldsymbol z(t)} g_{ik}(\boldsymbol z(t))\\ b_{ik2} &\equiv \mathop{\rm min}\limits_{\boldsymbol z(t)} g_{ik}(\boldsymbol z(t))\\ \end{aligned}\]

借助上述新定义的变量，可以将 $f_{ij}(\boldsymbol z(t)), g_{ik}(\boldsymbol z(t))$ 转化为用其最大最小值表达的形式（transforming into fuzzy model representation）：

\[\begin{aligned} f_{ij}(\boldsymbol z(t)) &= h_{ij1}(\boldsymbol z(t))a_{ij1} + h_{ij2}(\boldsymbol z(t))a_{ij2}\\ g_{ik}(\boldsymbol z(t)) &= v_{ik1}(\boldsymbol z(t))b_{ik1} + v_{ik2}(\boldsymbol z(t))b_{ik2} \end{aligned}\]

上式的权重参数满足

\[\sum_{l=1}^2 h_{ijl}(\boldsymbol z(t)) = 1\quad \sum_{l=1}^2 v_{ikl}(\boldsymbol z(t)) = 1\]

权重参数实质上就是隶属度函数的形式，可以定义如下

\[\begin{aligned} h_{ij1}(\boldsymbol z(t)) &= \frac{f_{ij}(\boldsymbol z(t))-a_{ij2}}{a_{ij1}-a_{ij2}}\\ h_{ij2}(\boldsymbol z(t)) &= \frac{a_{ij1} - f_{ij}(\boldsymbol z(t))}{a_{ij1}-a_{ij2}}\\ v_{ik1}(\boldsymbol z(t)) &= \frac{g_{ik}(\boldsymbol z(t))-b_{ij2}}{b_{ij1}-b_{ij2}}\\ v_{ik2}(\boldsymbol z(t)) &= \frac{b_{ik1} - g_{ik}(\boldsymbol z(t))}{b_{ik1}-b_{ik2}}\\ \end{aligned}\]

最终可以将原始TS模糊系统表示为：

\[\begin{aligned} \dot x_i(t) &= \sum_{j=1}^n f_{ij}(\boldsymbol z(t))x_j(t) + \sum_{k=1}^m g_{ik}(\boldsymbol z(t))u_k(t)\\ &=\sum_{j=1}^n\sum_{l=1}^2 h_{ijl}(\boldsymbol z(t))a_{ijl}x_j(t) + \sum_{k=1}^m\sum_{l=1}^2 v_{ikl}(\boldsymbol z(t))b_{ikl}u_k(t)\\ &=\sum_{l=1}^2 \left[ \begin{bmatrix} h_{i1l}a_{i1l}&\cdots&h_{inl}a_{inl} \end{bmatrix} \begin{bmatrix} x_1(t)\\ \vdots\\ x_n(t) \end{bmatrix}+ \begin{bmatrix} v_{i1l}b_{i1l}&\cdots&v_{iml}b_{iml} \end{bmatrix} \begin{bmatrix} u_1(t)\\ \vdots\\ u_m(t) \end{bmatrix} \right] \end{aligned}\]

$i$ 是输入向量的维度（表示状态方程的每个状态量），$j$ 也是输入向量的维度（表示每个状态量的一阶导与所有状态量的关系），$l$ 是取大取小值的维度。

将上述式子转为矩阵形式，如下

\[\begin{aligned} \dot \boldsymbol x(t) &=\sum_{i=1}^n\sum_{j=1}^n\sum_{l=1}^2 h_{ijl}(\boldsymbol z(t))a_{ijl} \boldsymbol U^A_{ij} \boldsymbol x(t) + \sum_{i=1}^n\sum_{k=1}^m\sum_{l=1}^2 v_{ikl}(\boldsymbol z(t))b_{ikl}\boldsymbol U^B_{ik}\boldsymbol u(t)\\ &=\sum_{i=1}^n\sum_{j=1}^n\sum_{l=1}^2 h_{ijl}(\boldsymbol z(t)) \boldsymbol A_{ijl} \boldsymbol x(t) + \sum_{i=1}^n\sum_{k=1}^m\sum_{l=1}^2 v_{ikl}(\boldsymbol z(t))\boldsymbol B_{ikl}\boldsymbol u(t)\\ \end{aligned}\]

其中

\[\begin{aligned} &\quad \quad \quad \quad \quad \quad \quad \quad j\\ \boldsymbol A_{ijl} &= i\begin{bmatrix} 0&\cdots&0&0&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0&\cdots&0&0&0&\cdots&0\\ 0&\cdots&0&a_{ijl}&0&\cdots&0\\ 0&\cdots&0&0&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0&\cdots&0&0&0&\cdots&0\\ \end{bmatrix}\\ &\quad \quad \quad \quad \quad \quad \quad \quad k\\ \boldsymbol B_{ikl} &= i\begin{bmatrix} 0&\cdots&0&0&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0&\cdots&0&0&0&\cdots&0\\ 0&\cdots&0&b_{ikl}&0&\cdots&0\\ 0&\cdots&0&0&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0&\cdots&0&0&0&\cdots&0\\ \end{bmatrix} \end{aligned}\]

个人理解：上述两个矩阵就是为了前面的矩阵求和式而人工构造的，可以与后面的 $\boldsymbol x(t), \boldsymbol u(t)$ 做矩阵乘法，取到对应的元素乘积。也就是说，上述两个矩阵分别有 $n^2,m^2$ 个，且每个都是上面这种只有一个元素不为 0 的稀疏形式。即

\[\begin{aligned} \dot \boldsymbol x(t) &=\sum_{l=1}^2 \left[ a_{11l}\begin{bmatrix} h_{11l}&\boldsymbol 0\\ \boldsymbol 0&\boldsymbol 0\\ \end{bmatrix}\boldsymbol x(t)+\cdots+ a_{nnl}\begin{bmatrix} \boldsymbol 0&\boldsymbol 0\\ \boldsymbol 0&h_{nnl}\\ \end{bmatrix}\boldsymbol x(t)+ b_{11l}\begin{bmatrix} h_{11l}&\boldsymbol 0\\ \boldsymbol 0&\boldsymbol 0\\ \end{bmatrix}\boldsymbol u(t)+\cdots+ b_{nnl}\begin{bmatrix} \boldsymbol 0&\boldsymbol 0\\ \boldsymbol 0&v_{nml}\\ \end{bmatrix}\boldsymbol u(t) \right]\\ &=\sum_{l=1}^2 \left[ \begin{bmatrix} h_{11l}a_{11l}&\cdots&h_{1nl}a_{1nl}\\ \vdots&\ddots&\vdots\\ h_{n1l}a_{n1l}&\cdots&h_{nnl}a_{nnl}\\ \end{bmatrix} \begin{bmatrix} x_1(t)\\ \vdots\\ x_n(t) \end{bmatrix}+ \begin{bmatrix} v_{11l}b_{11l}&\cdots&v_{1ml}b_{1ml}\\ \vdots&\ddots&\vdots\\ v_{n1l}b_{n1l}&\cdots&v_{nml}b_{nml}\\ \end{bmatrix} \begin{bmatrix} u_1(t)\\ \vdots\\ u_m(t) \end{bmatrix} \right]\\ &=\sum_{l=1}^2 \left[ \boldsymbol h_l*\boldsymbol A_l\cdot \boldsymbol x(t) + \boldsymbol v_l*\boldsymbol B_l\cdot \boldsymbol u(t) \right] \end{aligned}\]

作者表明，$a_{ijl}, b_{ikl}$ 再规则约减中非常重要，上面矩阵和的式子在规则约减中十分方便。

3.1.2. 模糊化表示

下面分析一般系统状态方程和广义 TS 模型之间的等价性，也即分析一般的系统状态方程怎么转化为广义 TS 模糊模型的形式。

首先给出结论

\[\begin{aligned} \dot \boldsymbol x(t) &=\sum_{i=1}^n\sum_{j=1}^n\sum_{l=1}^2 h_{ijl}(\boldsymbol z(t))a_{ijl} \boldsymbol U^A_{ij} \boldsymbol x(t)+\sum_{i=1}^n\sum_{k=1}^m\sum_{l=1}^2 v_{ikl}(\boldsymbol z(t))b_{ikl}\boldsymbol U^B_{ik}\boldsymbol u(t)\\ &=\sum_{p=1}^{2^{n(n+m)}} \hat h_p(\boldsymbol z(t))[\hat \boldsymbol A_p\boldsymbol x(t) + \hat \boldsymbol B_p\boldsymbol u(t)] \end{aligned}\]

下面进行一步步推导分析（原文中又是一个 where 易得，我人傻了）。

利用各项系数和为 1 的性质，进行连乘展开

\[\begin{aligned} 1 = &\overbrace{(h_{111}+h_{112})\cdots (h_{1n1}+h_{1n2})}^{n}&<1>\\ &\cdot(h_{211}+h_{212})\cdots (h_{2n1}+h_{2n2})&<2>\\ &\cdot\quad \cdots&\cdots\\ &\cdot(h_{n11}+h_{n12})\cdots (h_{nn1}+h_{nn2})&<n>\\ &\cdot\overbrace{(v_{111}+v_{n12})\cdots (v_{1m1}+v_{1m2})}^{m}&<1>\\ &\cdot\quad \cdots&\cdots\\ &\cdot(v_{n11}+v_{n12})\cdots (v_{nm1}+v_{nm2})&<n>\\ \end{aligned}\]

上式中一共有 $n\cdot n+n\cdot m$ 个括号，每个括号的和均为 1。下面从每个括号中任意取一个元素（$l=1\ or\ 2$）组成连乘项

  • 前 $n$ 行中，第一行展开后共有 $2^n$ 项，则前 $n$ 行一共有 $2^{n\cdot n}$ 项；
  • 后 $n$ 行中，第一行展开后共有 $2^m$ 项，则后 $n$ 行一共有 $2^{m\cdot n}$ 项。

那么，整个式子一共有 $C_{2^{n\cdot n}}^1C_{2^{m\cdot n}}^1=2^{n(n+m)}$ 项。每一项都是所有 $i,j,k$ 对不同 $l$ 的排列组合，即一共有 $2^{n(n+m)}$ 种排列组合。

假设选取所有括号里的 $l=1$（取所有括号里左边的元素），设该连乘项为第 $p=1$ 项， 则该项为

\[\begin{aligned} t_{p=1} = &(h_{111}\cdots h_{1n1})\cdots(h_{n11}\cdots h_{nn1})\\ &\cdot(v_{111}\cdots v_{1n1})\cdots(v_{1m1}\cdots v_{nm1})\\ = &\prod_{i=1}^n (h_{i11}\cdots h_{in1}) \cdot (v_{i11} \cdots v_{im1})\\ = &\prod_{i=1}^n \prod_{j=1}^n h_{ij1}\cdot (v_{i11}\cdots v_{im1})\\ = &\prod_{i=1}^n\prod_{j=1}^n\prod_{k=1}^m h_{ij1}v_{ik1} \end{aligned}\]

对所有的 $p$ 个连乘项求和，得到原始等式的最终表达形式

\[1 =\sum_{p=1}^{2^{n(n+m)}} t_p = \sum_{p=1}^{2^{n(n+m)}} \prod_{i=1}^n\prod_{j=1}^n\prod_{k=1}^m h_{ijl}v_{ikl}\]

其中 $l$ 与具体每项有关。

那么

\[\begin{aligned} \dot \boldsymbol x(t) &=\sum_{i=1}^n\sum_{j=1}^n\sum_{l=1}^2 h_{ijl}a_{ijl}\boldsymbol U_{ij}^A\boldsymbol x(t) + \sum_{i=1}^n\sum_{k=1}^m\sum_{l=1}^2 v_{ijl}(\boldsymbol z(t))b_{ikl}\boldsymbol U^B_{ik}\boldsymbol u(t)\\ &= \sum_{i=1}^n\sum_{j=1}^n\sum_{l=1}^2h_{ijl}\boldsymbol A_{ijl}\boldsymbol x(t) + \sum_{i=1}^n\sum_{k=1}^m\sum_{l=1}^2v_{ikl}\boldsymbol B_{ikl}\boldsymbol u(t)\\ &=\sum_{p=1}^{2^{n(n+m)}} \prod_{i=1}^n\prod_{j=1}^n\prod_{k=1}^m h_{ijl}v_{ikl} \left[ \sum_{i=1}^n\sum_{j=1}^n\sum_{l=1}^2h_{ijl}\boldsymbol A_{ijl}\boldsymbol x(t) + \sum_{i=1}^n\sum_{k=1}^m\sum_{l=1}^2v_{ikl}\boldsymbol B_{ikl}\boldsymbol u(t)\right] \\ \end{aligned}\]

然后我就推不出来了！

3.1.3. 举例

考虑如下非线性系统

\[\begin{aligned} \left\{ \begin{array}{l} \dot x_1(t) = x_2(t)\\ \dot x_2(t) = x_1(t){\rm cos}x_2(t)-x_3(t)\\ \dot x_3(t) = x_1(t)x_3(t) + (1+\varepsilon {\rm sin}x_3(t))u(t) \end{array} \right. \end{aligned}\]

状态量的取值范围为

\[\begin{aligned} \underline d_1\leq x_1(t)\leq \overline d_1\\ \underline d_2\leq x_2(t)\leq \overline d_2\\ \underline d_3\leq x_3(t)\leq \overline d_3\\ \end{aligned}\]

其中

\[\begin{aligned} \underline d_1 &= -5\\ \overline d_1 &= 5\\ \underline d_2 &= -\pi/2\\ \overline d_2 &= \pi/2\\ \underline d_3 &= -\pi\\ \overline d_3 &= \pi\\ \varepsilon &= 0.5\\ \end{aligned}\]

根据前面的约定，重新整理状态方程如下

\[\begin{aligned} \left\{ \begin{array}{l} \dot x_1(t) &= 0\cdot x_1(t)& + 1\cdot x_2(t)& + 0\cdot x_3(t)& + 0\cdot u(t)&\\ \dot x_2(t) &= {\rm cos}x_2(t)\cdot x_1(t)& + 0\cdot x_2(t)& +(-1) \cdot x_3(t)& + 0\cdot u(t)&\\ \dot x_3(t) &= 0\cdot x_1(t)& +0\cdot x_2(t)& + x_1(t)\cdot x_3(t)& + (1+\varepsilon {\rm sin}x_3(t))u(t)& \end{array} \right. \end{aligned}\]

有

\[\begin{aligned} f_{11}(z(t)) &= 0,\ &f_{12}(z(t)) = 1,\ &f_{13}(z(t)) = 0,\ &g_{11}(z(t)) = 0\\ f_{21}(z(t)) &= {\rm cos}x_2(t),\ &f_{22}(z(t)) = 0,\ &f_{23}(z(t)) = -1,\ &g_{11}(z(t)) = 0\\ f_{31}(z(t)) &= 0,\ &f_{32}(z(t)) = 0,\ &f_{33}(z(t)) = x_1(t),\ &g_{11}(z(t)) = (1+\varepsilon {\rm sin}x_3(t))\\ \end{aligned}\]

计算系数的最大最小值，有（加 * 号的是原文中列出的，略去了 $a=b=0$ 项）

\[\begin{aligned} a_{111} &= 0,\ &a_{112} = 0\\ *a_{121} &= 1,\ &a_{122} = 1\\ a_{131} &= 0,\ &a_{132} = 0\\ b_{111} &= 0,\ &b_{112} = 0\\ \\ *a_{211} &= 1,\ &a_{212} = 0\\ a_{221} &= 0,\ &a_{222} = 0\\ *a_{231} &= -1,\ &a_{232} = -1\\ b_{211} &= 0,\ &b_{212} = 0\\ \\ a_{311} &= 0,\ &a_{312} = 0\\ a_{321} &= 0,\ &a_{322} = 0\\ *a_{331} &= 5,\ &a_{332} = -5\\ *b_{311} &= 1.5,\ &b_{312} = 0.5\\ \end{aligned}\]

隶属度函数指定为（加 * 号的是原文中列出的，略去了 $a=b=0$ 对应项）

\[\begin{aligned} h_{111}(z(t)) &= 0.5,\ &h_{112}(z(t)) = 0.5\\ *h_{121}(z(t)) &= 0.5,\ &h_{122}(z(t)) = 0.5\\ h_{131}(z(t)) &= 0.5,\ &h_{132}(z(t)) = 0.5\\ v_{111}(z(t)) &= 0.5,\ &v_{112}(z(t)) = 0.5\\ \\ *h_{211}(z(t)) &= {\rm cos}x_2(t),\ &h_{212}(z(t)) = 1-{\rm cos}x_2(t)\\ h_{221}(z(t)) &= 0.5,\ &h_{222}(z(t)) = 0.5\\ *h_{231}(z(t)) &= 0.5,\ &h_{232}(z(t)) = 0.5\\ v_{211}(z(t)) &= 0.5,\ &v_{212}(z(t)) = 0.5\\ \\ h_{311}(z(t)) &= 0.5,\ &h_{312}(z(t)) = 0.5\\ h_{321}(z(t)) &= 0.5,\ &h_{322}(z(t)) = 0.5\\ *h_{331}(z(t)) &= \frac{x_1(t)+5}{10},\ &h_{332}(z(t)) = \frac{5-x_1(t)}{10}\\ *v_{311}(z(t)) &= \frac{1+{\rm sin}x_3(t)}{2},\ &v_{312}(z(t)) = \frac{1-{\rm sin}x_3(t)}{2}\\ \end{aligned}\]

根据一般形式

\[\dot \boldsymbol x(t) =\sum_{i=1}^n\sum_{j=1}^n\sum_{l=1}^2 h_{ijl}(\boldsymbol z(t))a_{ijl} \boldsymbol U^A_{ij} \boldsymbol x(t) + \sum_{i=1}^n\sum_{k=1}^m\sum_{l=1}^2 v_{ikl}(\boldsymbol z(t))b_{ikl}\boldsymbol U^B_{ik}\boldsymbol u(t)\]

则原系统对应的模糊模型可以写为（所有 0 项乘积忽略）

\[\begin{aligned} \dot \boldsymbol x(t) = &h_{121}a_{121}\boldsymbol U_{12}^A \boldsymbol x(t)+h_{122}a_{122}\boldsymbol U_{12}^A \boldsymbol x(t)+\\ & h_{211}a_{211}\boldsymbol U^A_{21} \boldsymbol x(t)+h_{212}a_{212}\boldsymbol U^A_{21} \boldsymbol x(t)+\\ & h_{231}a_{231}\boldsymbol U^A_{23} \boldsymbol x(t)+h_{232}a_{232}\boldsymbol U^A_{23} \boldsymbol x(t)+\\ & h_{331}a_{331}\boldsymbol U^A_{33} \boldsymbol x(t)+h_{332}a_{332}\boldsymbol U^A_{33} \boldsymbol x(t)+\\ & v_{311}b_{311}\boldsymbol U^B_{31} \boldsymbol u(t)+v_{312}b_{312}\boldsymbol U^B_{31} \boldsymbol u(t) \end{aligned}\]

进行矩阵化，$h, a$ 的前两个下标以及 $U$ 的下标为行列号，$h, a$ 的第三个下标为最大最小值的加权和作为元素值，列写出子矩阵

\[\begin{aligned} \boldsymbol A_1 &= h_{121}a_{121}\boldsymbol U_{12}^A\\ \boldsymbol A_2 &= h_{122}a_{122}\boldsymbol U_{12}^A\\ \boldsymbol A_3 &= h_{211}a_{211}\boldsymbol U_{21}^A\\ \boldsymbol A_4 &= h_{212}a_{212}\boldsymbol U_{21}^A\\ \boldsymbol A_5 &= h_{231}a_{231}\boldsymbol U_{23}^A\\ \boldsymbol A_6 &= h_{232}a_{232}\boldsymbol U_{23}^A\\ \boldsymbol A_7 &= h_{331}a_{331}\boldsymbol U_{33}^A\\ \boldsymbol A_8 &= h_{332}a_{332}\boldsymbol U_{33}^A\\ \boldsymbol B_1 &= v_{311}a_{311}\boldsymbol U_{31}^B\\ \boldsymbol B_2 &= v_{312}a_{312}\boldsymbol U_{31}^B\\ \end{aligned}\]

令

\[\begin{aligned} \boldsymbol A &= \boldsymbol A_1+\boldsymbol A_2+\cdots+\boldsymbol A_8\\ \boldsymbol B &= \boldsymbol B_1+\boldsymbol B_2 \end{aligned}\]

有

\[\begin{aligned} \dot \boldsymbol x(t) &= \boldsymbol A\boldsymbol x(t) + \boldsymbol B\boldsymbol u(t)\\ \end{aligned}\]

但是这并不是广义模糊 TS 系统的形式，下面进行转换。

注意到（存在两组最大最小值相等的情况）

\[a_{121}=a_{122}=1,\ a_{231}=a_{232}=-1\]

则

\[\begin{aligned} \boldsymbol A_1 &= \boldsymbol A_2 = 0.5\boldsymbol U_{12}^A\\ \boldsymbol A_5 &= \boldsymbol A_6 = -0.5\boldsymbol U_{23}^A\\ \boldsymbol A_{b1} &= \boldsymbol A_1 + \boldsymbol A_2 = \begin{bmatrix} 0&1&0\\ 0&0&0\\ 0&0&0 \end{bmatrix}\\ \boldsymbol A_{b2} &= \boldsymbol A_5 + \boldsymbol A_6 = \begin{bmatrix} 0&0&0\\ 0&0&-1\\ 0&0&0 \end{bmatrix}\\ \boldsymbol A_{b} &= \boldsymbol A_{b1} + \boldsymbol A_{b2} = \begin{bmatrix} 0&1&0\\ 0&0&-1\\ 0&0&0 \end{bmatrix} \end{aligned}\]

分析其它子矩阵。将子矩阵进行变换，如下

\[\begin{aligned} \boldsymbol A_3 &= h_{211}a_{211}\boldsymbol U_{21}^A = h_{211}(v_{311}+v_{312})(h_{331}+h_{332})a_{211}\boldsymbol U_{21}^A\\ \boldsymbol A_4 &= h_{212}a_{211}\boldsymbol U_{21}^A = h_{212}(v_{311}+v_{312})(h_{331}+h_{332})a_{212}\boldsymbol U_{21}^A\\ \boldsymbol A_7 &= h_{331}a_{331}\boldsymbol U_{33}^A = h_{331}(v_{311}+v_{312})(h_{211}+h_{212}) a_{331}\boldsymbol U_{33}^A\\ \boldsymbol A_8 &= h_{332}a_{332}\boldsymbol U_{33}^A = h_{332}(v_{311}+v_{312})(h_{211}+h_{212})a_{332}\boldsymbol U_{33}^A\\ \boldsymbol B_1 &= v_{311}b_{311}\boldsymbol U_{31}^B= v_{311}(h_{211}+h_{212})(h_{331}+h_{332})b_{311}\boldsymbol U_{31}^B\\ \boldsymbol B_2 &= v_{312}b_{312}\boldsymbol U_{31}^B= v_{312}(h_{211}+h_{212})(h_{331}+h_{332})b_{312}\boldsymbol U_{31}^B\\ \end{aligned}\]

类似的，将前面的两个矩阵的系数进行配合

\[\begin{aligned} \boldsymbol A_{b} &= (h_{211}+h_{212})(v_{311}+v_{312})(h_{331}+h_{332})\boldsymbol A_{b}\\ \end{aligned}\]

将系数展开后得到 八 组系数

\[\begin{aligned} h_1 &= h_{211}h_{331}v_{311}\\ h_2 &= h_{212}h_{331}v_{311}\\ h_3 &= h_{211}h_{332}v_{311}\\ h_4 &= h_{212}h_{332}v_{311}\\ h_5 &= h_{211}h_{331}v_{312}\\ h_6 &= h_{212}h_{331}v_{312}\\ h_7 &= h_{211}h_{332}v_{312}\\ h_8 &= h_{212}h_{332}v_{312}\\ \end{aligned}\]

则

\[\begin{aligned} \boldsymbol A_{b} &= (h_1+h_2+h_3+h_4+h_5+h_6+h_7+h_8)\boldsymbol A_{b}\\ \boldsymbol A_3 &=(h_1+h_3+h_5+h_7)a_{211}\boldsymbol U_{21}^A\\ \boldsymbol A_4 &=(h_2+h_4+h_6+h_8)a_{212}\boldsymbol U_{21}^A\\ \boldsymbol A_7 &=(h_1+h_2+h_5+h_6)a_{331}\boldsymbol U_{33}^A\\ \boldsymbol A_8 &=(h_3+h_4+h_7+h_8)a_{332}\boldsymbol U_{33}^A\\ \boldsymbol B_1 &= (h_1+h_2+h_3+h_4)\boldsymbol b_{311}U_{31}^B\\ \boldsymbol B_2 &= (h_5+h_6+h_7+h_8)\boldsymbol b_{312}U_{31}^B\\ \end{aligned}\]

按照系数 $h$ 重新整理矩阵，有

\[\begin{aligned} \boldsymbol A = &h_1\boldsymbol A_1+h_2\boldsymbol A_2+\cdots+h_8\boldsymbol A_8 +\\ &h_1\boldsymbol B_1 + h_2\boldsymbol B_2 + \cdots + h_8\boldsymbol B_8 \end{aligned}\]

其中

\[\begin{aligned} \boldsymbol A_1 &= \boldsymbol A_5 = \boldsymbol A_b + a_{211}\boldsymbol U_{21}^A + a_{331}\boldsymbol U_{33}^A = \begin{bmatrix} 0&1&0\\ \boldsymbol 1&0&-1\\ 0&0&\boldsymbol 5 \end{bmatrix}\\ \boldsymbol A_2 &= \boldsymbol A_6 = \boldsymbol A_b + a_{212}\boldsymbol U_{21}^A + a_{331}\boldsymbol U_{33}^A = \begin{bmatrix} 0&1&0\\ \boldsymbol 0&0&-1\\ 0&0&\boldsymbol 5 \end{bmatrix}\\ \boldsymbol A_3 &= \boldsymbol A_7 = \boldsymbol A_b + a_{211}\boldsymbol U_{21}^A + a_{332}\boldsymbol U_{33}^A = \begin{bmatrix} 0&1&0\\ \boldsymbol 1&0&-1\\ 0&0&-\boldsymbol 5 \end{bmatrix}\\ \boldsymbol A_4 &= \boldsymbol A_8 = \boldsymbol A_b + a_{212}\boldsymbol U_{21}^A + a_{332}\boldsymbol U_{33}^A = \begin{bmatrix} 0&1&0\\ \boldsymbol 0&0&-1\\ 0&0&-\boldsymbol 5 \end{bmatrix}\\ \boldsymbol B_1 &= \boldsymbol B_2 = \boldsymbol B_3 = \boldsymbol B_4 = b_{311}U_{31}^B = \begin{bmatrix} 0\\ 0\\ 1.5 \end{bmatrix}\\ \boldsymbol B_5 &= \boldsymbol B_6 = \boldsymbol B_7 = \boldsymbol B_8 = b_{312}U_{31}^B = \begin{bmatrix} 0\\ 0\\ 0.5 \end{bmatrix}\\ \end{aligned}\]

至此终于推得文中 「易得」 的系数，汗！

按照整理后的系数和矩阵重新写系统状态方程，有

\[\dot \boldsymbol x(t) = \sum_{i=1}^8 h_i(\boldsymbol z(t))[\boldsymbol A_i\boldsymbol x(t) + \boldsymbol B_i\boldsymbol u(t)]\]

Note that this fuzzy model has nonlinear terms in $A(2,1),A(3,3)$, and $B(3,1)$, where denotes the (2, 1) element of $A$ matrix.

原文说模糊系统的 $\boldsymbol A$ 的 $(2,1),(3,3)$ 元素和矩阵 $\boldsymbol B$ 的 $(3,1)$ 元素是非线性项。个人觉得，单纯从系数矩阵而言 $\boldsymbol A(3,3)$ 并不是非线性项，但是对于整个系统而言的确是非线性的。

3.2. 规则约减

3.2.1. 约减方式

规则约减与使用 LMI 进行控制器设计的计算工作量密切相关。

基本思路：将非线性项 $f_{ij}(\boldsymbol z(t)), g_{ik}(\boldsymbol z(t))$ 替换为常数项 $a_{i_0j_0},b_{i_0k_0}$，其中 $a_{i_0j_0}=(a_{ij1}+a_{ij2})/2,\ b_{i_0k_0} = (b_{ik1}+b_{ik2})/2$。

对于任意的 $i_0,j_0$，对 $f_{i_0j_0}(\boldsymbol z(t))$ 约减后的模型为

\[\begin{aligned} \dot \boldsymbol x(t) = &\mathop{\sum_{i=1}^n\sum_{j=1}^n\sum_{l=1}^2}\limits_{(i,j)\neq (i_0,j_0)} h_{ijl}(\boldsymbol z(t))a_{ijl} \boldsymbol U^A_{ij} \boldsymbol x(t)\\ &+a_{i_0j_0}\boldsymbol U^A_{i_0j_0} \boldsymbol x(t)\\ &+\sum_{i=1}^n\sum_{k=1}^m\sum_{l=1}^2 v_{ijl}(\boldsymbol z(t))b_{ikl}\boldsymbol U^B_{ik}\boldsymbol u(t) \end{aligned}\]

类似地，对于任意的 $i_0,k_0$，对 $g_{i_0j_0}(\boldsymbol z(t))$ 约减后的模型为

\[\begin{aligned} \dot \boldsymbol x(t) = &\sum_{i=1}^n\sum_{j=1}^n\sum_{l=1}^2 h_{ijl}(\boldsymbol z(t))a_{ijl} \boldsymbol U^A_{ij} \boldsymbol x(t)\\ &+\mathop{\sum_{i=1}^n\sum_{k=1}^m\sum_{l=1}^2}\limits_{(i,k)\neq (i_0,k_0)} v_{ikl}(\boldsymbol z(t))b_{ikl}\boldsymbol U^B_{ik}\boldsymbol u(t)\\ &+b_{i_0k_0}\boldsymbol U^B_{i_0k_0} \boldsymbol u(t) \end{aligned}\]

个人理解，假设针对第 $(i,j)=(1,1)$ 个 $f_{ij}$ 进行规则约减，有

\[\begin{aligned} \dot \boldsymbol x(t) &=\sum_{l=1}^2 \left[ \begin{bmatrix} h_{11l}a_{11l}&\cdots&h_{1nl}a_{1nl}\\ \vdots&\ddots&\vdots\\ h_{n1l}a_{n1l}&\cdots&h_{nnl}a_{nnl}\\ \end{bmatrix} \begin{bmatrix} x_1(t)\\ \vdots\\ x_n(t) \end{bmatrix}+ \begin{bmatrix} v_{11l}b_{11l}&\cdots&v_{1ml}b_{1ml}\\ \vdots&\ddots&\vdots\\ v_{n1l}b_{n1l}&\cdots&v_{nml}b_{nml}\\ \end{bmatrix} \begin{bmatrix} u_1(t)\\ \vdots\\ u_m(t) \end{bmatrix} \right]\\ &=\begin{bmatrix} h_{111}a_{111}+h_{112}a_{112}&\cdots&h_{1n1}a_{1n1}+h_{1n2}a_{1n2}\\ \vdots&\ddots&\vdots\\ h_{n11}a_{n11}+h_{n12}a_{n12}&\cdots&h_{nn1}a_{nn1}+h_{nn2}a_{nn2}\\ \end{bmatrix} \begin{bmatrix} x_1(t)\\ \vdots\\ x_n(t) \end{bmatrix}+\cdots\\ &=\begin{bmatrix} (a_{111}+a_{112})/2&\cdots&h_{1n1}a_{1n1}+h_{1n2}a_{1n2}\\ \vdots&\ddots&\vdots\\ h_{n11}a_{n11}+h_{n12}a_{n12}&\cdots&h_{nn1}a_{nn1}+h_{nn2}a_{nn2}\\ \end{bmatrix} \begin{bmatrix} x_1(t)\\ \vdots\\ x_n(t) \end{bmatrix}+\cdots \end{aligned}\]

同理，对第 $(i,k)=(1,1)$ 个 $g_{ik}$ 进行规则约减，有

\[\begin{aligned} \dot \boldsymbol x(t) &=\cdots + \begin{bmatrix} (b_{111}+b_{112})/2&\cdots&v_{1m1}b_{1m1}+v_{1m2}v_{1m2}\\ \vdots&\ddots&\vdots\\ v_{n11}b_{n11}+v_{n12}b_{n12}&\cdots&v_{nm1}b_{nm1}+v_{nm2}b_{nm2}\\ \end{bmatrix} \begin{bmatrix} u_1(t)\\ \vdots\\ u_m(t) \end{bmatrix} \end{aligned}\]

3.2.2. 模型不确定性

对原始模型进行规则约减后，会存在约减偏差，作者将其转化为模型不确定性。

  • 任一约减

对于某一 $i_0,j_0,k_0$ 项进行规则约减，假设模型不确定性为 $\delta^A_{i_0j_0}(t),\delta^B_{i_0k_0}(t)$，已知

\[a_{i_0j_0}=\frac{a_{ij1}+a_{ij2}}{2},\ b_{i_0k_0}=\frac{b_{ik1}+b_{ik2}}{2}\]

那么原模型可写为

\[\begin{aligned} \dot \boldsymbol x(t) = &\mathop{\sum_{i=1}^n\sum_{j=1}^n\sum_{l=1}^2}\limits_{(i,j)\neq (i_0,j_0)} h_{ijl}(\boldsymbol z(t))a_{ijl} \boldsymbol U^A_{ij} \boldsymbol x(t)\\ &+(a_{i_0j_0} + \delta^A_{i_0j_0}(t))\boldsymbol U^A_{i_0j_0} \boldsymbol x(t)\\ &+\mathop{\sum_{i=1}^n\sum_{k=1}^m\sum_{l=1}^2}\limits_{(i,k)\neq (i_0,k_0)} v_{ikl}(\boldsymbol z(t))b_{ikl}\boldsymbol U^B_{ik}\boldsymbol u(t)\\ &+(b_{i_0k_0}+\delta^B_{i_0k_0}(t))\boldsymbol U^B_{i_0k_0} \boldsymbol u(t) \end{aligned}\]

这种约减导致的偏差，最大不会超过对应非线性项取值范围的一半（比如该项实际取值为最小值，结果我们用其平均值作为替代，此时偏差正好为取值范围的一半）。即

\[\vert\vert \delta^A_{i_0j_0}(t) \vert\vert \leq \frac{a_{ij1}-a_{ij2}}{2},\ \vert\vert \delta^B_{i_0k_0}(t) \vert\vert\leq \frac{b_{ik1}-b_{ik2}}{2}\]

对模糊化后的系统方程进行重新表达，令

\[\begin{aligned} r&=2^{n(n+m)} \end{aligned}\]

则系统可以写为

\[\begin{aligned} \dot \boldsymbol x(t) &=\sum_{p=1}^{2^{n(n+m)}} \hat h_p(\boldsymbol z(t))[\hat \boldsymbol A_p\boldsymbol x(t) + \hat \boldsymbol B_p\boldsymbol u(t)]\\ &= \sum_{p=1}^{\frac{1}{4}r}h_p(\boldsymbol z(t))[\boldsymbol A_p\boldsymbol x(t)+\boldsymbol B_p\boldsymbol u(t) +\delta^A_{i_0j_0}(t) \boldsymbol U_{i_0j_0}\boldsymbol x(t) +\delta^B_{i_0k_0}(t) \boldsymbol U_{i_0k_0}\boldsymbol u(t)]\\ &=\sum_{p=1}^{\frac{1}{4}r}h_p(\boldsymbol z(t))[ (\boldsymbol A_p + \boldsymbol D_{ap}\Delta_{ap}(t)\boldsymbol E_{ap})\boldsymbol x(t)+(\boldsymbol B_p+ \boldsymbol D_{bp}\Delta_{bp}(t)\boldsymbol E_{bp})\boldsymbol u(t)] \end{aligned}\]

其中，$\frac{1}{4}r$ 是因为系统中有两项（$f_{ij},g_{ik}$）被约减了。每少一项，需要遍历的参数减半（参考前面系数和的连乘的分析），因此总规则个数需要除以 $2\cdot 2=4$。$\boldsymbol A_p,\boldsymbol B_p$ 为去除 $(i_0,j_0)$ 元素后的系数矩阵。矩阵 $\boldsymbol D_{ap},\boldsymbol E_{ap},\boldsymbol D_{bp},\boldsymbol E_{bp}$ 为

\[\begin{aligned} \boldsymbol D_{ap} &= \boldsymbol D_{bp} = i_0\begin{bmatrix} 0\\\vdots\\0\\1\\0\\ \vdots\\ 0 \end{bmatrix}\\ &\quad \quad \quad \quad \ \ \ j_0\\ \boldsymbol E_{ap} &= [0\ \cdots\ 0\ 1\ 0\ \cdots\ 0]\\ &\quad \quad \quad \quad \ \ \ k_0\\ \boldsymbol E_{bp} &= [0\ \cdots\ 0\ 1\ 0\ \cdots\ 0]\\ \end{aligned}\]
  • 一般约减

对于任意 $i,j,k$ 项进行规则约减，引入模型不确定性 $\boldsymbol \Delta^A_{ij}(t),\boldsymbol \Delta^B_{ik}(t)$，有

\[\begin{aligned} \dot \boldsymbol x(t) &=\sum_{i=1}^n\sum_{j=1}^n (a_{ij}+\delta^A_{ij}(t))\boldsymbol U_{ij}^A\boldsymbol x(t) + \sum_{i=1}^n\sum_{k=1}^m (b_{ik}+\delta^B_{ik}(t))\boldsymbol U_{ik}^B\boldsymbol u(t)\\ \end{aligned}\]

其中

\[\vert\vert \delta^A_{ij}(t) \vert\vert \leq \frac{a_{ij1}-a_{ij2}}{2},\ \vert\vert \delta^B_{ik}(t) \vert\vert\leq \frac{b_{ik1}-b_{ik2}}{2}\]

写成矩阵的形式

\[\dot \boldsymbol x(t)=\sum_{i=1}^{\frac{1}{4}r} (\boldsymbol A_i+\boldsymbol D_{ai}\boldsymbol \Delta_{ai}(t)\boldsymbol E_{ai})\boldsymbol x(t)+(\boldsymbol B_i+ \boldsymbol D_{bi}\boldsymbol \Delta_{bi}(t)\boldsymbol E_{bi})\boldsymbol u(t)\]

其中

\[\vert\vert \delta^A_{ij}(t) \vert\vert \leq \frac{1}{\rho_{ai}},\ \vert\vert \delta^B_{ik}(t) \vert\vert\leq \frac{1}{\rho_{bi}}\]

$\frac{1}{\rho_{ai}},\frac{1}{\rho_{bi}}$ 根据前面不确定性不等式的上界决定（原文可能误写作 $\rho_{ai},\rho_{bi}$）。

  • 全部约减

对于所有 $i,j,k$ 项进行规则约减，那么很显然，规则最终从 $2^{n(n+m)}$ 个变成 1 个。

引入模型不确定性 $\delta^A_{ij}(t),\delta^B_{ik}(t)$，有

\[\begin{aligned} \dot \boldsymbol x(t) &=\sum_{i=1}^n\sum_{j=1}^n (a_{ij}+\delta^A_{ij}(t))\boldsymbol U_{ij}^A\boldsymbol x(t) + \sum_{i=1}^n\sum_{k=1}^m (b_{ik}+\delta^B_{ik}(t))\boldsymbol U_{ik}^B\boldsymbol u(t)\\ \end{aligned}\]

其中

\[\vert\vert \delta^A_{ij}(t) \vert\vert \leq \frac{a_{ij1}-a_{ij2}}{2},\ \vert\vert \delta^B_{ik}(t) \vert\vert\leq \frac{b_{ik1}-b_{ik2}}{2}\]

写成矩阵的形式

\[\dot \boldsymbol x(t)=(\boldsymbol A_1+\boldsymbol D_{a1}\boldsymbol \Delta_{a1}(t)\boldsymbol E_{a1})\boldsymbol x(t)+(\boldsymbol B_1+ \boldsymbol D_{b1}\boldsymbol \Delta_{b1}(t)\boldsymbol E_{b1})\boldsymbol u(t)\]

其中

\[\begin{aligned} \boldsymbol A_1 &= \begin{bmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\\ \end{bmatrix}\in \mathbb R^{n\times n},\ \boldsymbol B_1 = \begin{bmatrix} b_{11}&b_{12}&\cdots&a_{1m}\\ b_{21}&b_{22}&\cdots&a_{2m}\\ \vdots&\vdots&\ddots&\vdots\\ b_{n1}&b_{n2}&\cdots&a_{nm}\\ \end{bmatrix}\in \mathbb R^{n\times m}\\ \boldsymbol \Delta_{a1}(t) &= diag(\delta_{11}^A(t)\cdots\delta_{1n}^A(t)\cdots\delta_{nn}^A(t)) = \begin{bmatrix} \delta_{11}^A(t)&&\boldsymbol 0\\ &\ddots&\\ \boldsymbol 0&&\delta_{nn}^A(t) \end{bmatrix}\in R^{n^2\times n^2}\\ \boldsymbol \Delta_{b1}(t) &= diag(\delta_{11}^B(t)\cdots\delta_{1m}^B(t)\cdots\delta_{nm}^B(t)) = \begin{bmatrix} \delta_{11}^B(t)&&\boldsymbol 0\\ &\ddots&\\ \boldsymbol 0&&\delta_{nm}^B(t) \end{bmatrix}\in R^{nm\times nm}\\ \boldsymbol D_{a1} &= \begin{bmatrix} \overbrace{1 1\cdots 1}^n&&\boldsymbol 0\\ \vdots&\ddots&\vdots\\ \boldsymbol 0&&\overbrace{1 1\cdots 1}^n \end{bmatrix}\in \mathbb R^{n\times n^2},\ \boldsymbol D_{b1} = \begin{bmatrix} \overbrace{1 1\cdots 1}^m&&\boldsymbol 0\\ \vdots&\ddots&\vdots\\ \boldsymbol 0&&\overbrace{1 1\cdots 1}^m \end{bmatrix}\in \mathbb R^{n\times nm}\\ \boldsymbol E_{a1} &= \begin{bmatrix} \boldsymbol I_n\\ \vdots\\ \boldsymbol I_n\\ \end{bmatrix} =\begin{bmatrix} 1&&\boldsymbol 0\\ &\ddots&\\ \boldsymbol 0&&1\\ 1&&\boldsymbol 0\\ &\ddots&\\ \boldsymbol 0&&1\\ &\vdots&\\ \boldsymbol 0&&1\\ \end{bmatrix}\in \mathbb R^{n^2\times n},\ \boldsymbol E_{b1} = \begin{bmatrix} \boldsymbol I_m\\ \vdots\\ \boldsymbol I_m\\ \end{bmatrix} \in \mathbb R^{nm\times m}, \end{aligned}\]

偏差矩阵 $\boldsymbol D_{a1}\boldsymbol \Delta_{a1}(t)\boldsymbol E_{a1}$ 和系数矩阵 $\boldsymbol A_1$ 的形式一致，虽然我并不明白为什么非要这么写成这么复杂的矩阵形式

\[\begin{aligned} \boldsymbol D_{a1}\boldsymbol \Delta_{a1}(t)\boldsymbol E_{a1} &= \begin{bmatrix} \overbrace{1 1\cdots 1}^n&&\boldsymbol 0\\ \vdots&\ddots&\vdots\\ \boldsymbol 0&&\overbrace{1 1\cdots 1}^n \end{bmatrix}_{n\times n^2} \begin{bmatrix} \delta_{11}^A(t)&&\boldsymbol 0\\ &\ddots&\\ \boldsymbol 0&&\delta_{nn}^A(t) \end{bmatrix}_{n^2\times n^2} \begin{bmatrix} 1&&\boldsymbol 0\\ &\ddots&\\ \boldsymbol 0&&1\\ 1&&\boldsymbol 0\\ &\ddots&\\ \boldsymbol 0&&1\\ &\vdots&\\ \boldsymbol 0&&1\\ \end{bmatrix}_{n^2\times n}\\ &=\begin{bmatrix} \overbrace{\delta_{11}^A(t)\ \delta_{12}^A(t)\ \cdots \delta_{1n}^A(t)}^n&&\boldsymbol 0\\ \vdots&\ddots&\vdots\\ \boldsymbol 0&&\overbrace{\delta_{n1}^A(t)\ \delta_{n2}^A(t)\ \cdots \delta_{nn}^A(t)}^n \end{bmatrix}_{n\times n^2} \begin{bmatrix} 1&&\boldsymbol 0\\ &\ddots&\\ \boldsymbol 0&&1\\ 1&&\boldsymbol 0\\ &\ddots&\\ \boldsymbol 0&&1\\ &\vdots&\\ \boldsymbol 0&&1\\ \end{bmatrix}_{n^2\times n}\\ &=\begin{bmatrix} \delta_{11}^A(t)&\delta_{12}^A(t)&\cdots&\delta_{1n}^A(t)\\ \delta_{21}^A(t)&\delta_{22}^A(t)&\cdots&\delta_{2n}^A(t)\\ \vdots&\vdots&\ddots&\vdots\\ \delta_{n1}^A(t)&\delta_{n2}^A(t)&\cdots&\delta_{nn}^A(t)\\ \end{bmatrix} \end{aligned}\]

偏差矩阵 $\boldsymbol D_{b1}\boldsymbol \Delta_{b1}(t)\boldsymbol E_{b1}$ 和系数矩阵 $\boldsymbol B_1$ 类似，这里不再赘述。

3.2.3. 举例

对前面已经完成建模的系统（3.1.3. 举例）进行规则约减。系统状态方程为

\[\dot \boldsymbol x(t) = \sum_{i=1}^8 h_i(\boldsymbol z(t))[\boldsymbol A_i\boldsymbol x(t) + \boldsymbol B_i\boldsymbol u(t)]\]

其中

\[\begin{aligned} h_1 &= h_{211}h_{331}v_{311}={\rm cos}x_2(t)\cdot\frac{5+x_1(t)}{10}\cdot\frac{1+{\rm sin}x_3(t)}{2}\\ h_2 &= h_{212}h_{331}v_{311}=(1-{\rm cos}x_2(t))\cdot\frac{5+x_1(t)}{10}\cdot\frac{1+{\rm sin}x_3(t)}{2}\\ h_3 &= h_{211}h_{332}v_{311}={\rm cos}x_2(t)\cdot\frac{5-x_1(t)}{10}\cdot\frac{1+{\rm sin}x_3(t)}{2}\\ h_4 &= h_{212}h_{332}v_{311}=(1-{\rm cos}x_2(t))\cdot\frac{5-x_1(t)}{10}\cdot\frac{1+{\rm sin}x_3(t)}{2}\\ h_5 &= h_{211}h_{331}v_{312}={\rm cos}x_2(t)\cdot\frac{5+x_1(t)}{10}\cdot\frac{1-{\rm sin}x_3(t)}{2}\\ h_6 &= h_{212}h_{331}v_{312}=(1-{\rm cos}x_2(t))\cdot\frac{5+x_1(t)}{10}\cdot\frac{1-{\rm sin}x_3(t)}{2}\\ h_7 &= h_{211}h_{332}v_{312}={\rm cos}x_2(t)\cdot\frac{5-x_1(t)}{10}\cdot\frac{1-{\rm sin}x_3(t)}{2}\\ h_8 &= h_{212}h_{332}v_{312}=(1-{\rm cos}x_2(t))\cdot\frac{5-x_1(t)}{10}\cdot\frac{1-{\rm sin}x_3(t)}{2}\\ \boldsymbol A_1 &= \boldsymbol A_5 = \begin{bmatrix} 0&1&0\\ \boldsymbol 1&0&-1\\ 0&0&\boldsymbol 5 \end{bmatrix},\ \boldsymbol A_2 = \boldsymbol A_6 = \begin{bmatrix} 0&1&0\\ \boldsymbol 0&0&-1\\ 0&0&\boldsymbol 5 \end{bmatrix}\\ \boldsymbol A_3 &= \boldsymbol A_7 = \begin{bmatrix} 0&1&0\\ \boldsymbol 1&0&-1\\ 0&0&-\boldsymbol 5 \end{bmatrix},\ \boldsymbol A_4 = \boldsymbol A_8 = \begin{bmatrix} 0&1&0\\ \boldsymbol 0&0&-1\\ 0&0&-\boldsymbol 5 \end{bmatrix}\\ \boldsymbol B_1 &= \boldsymbol B_2 = \boldsymbol B_3 = \boldsymbol B_4 = \begin{bmatrix} 0\\ 0\\ 1.5 \end{bmatrix},\ \boldsymbol B_5 = \boldsymbol B_6 = \boldsymbol B_7 = \boldsymbol B_8 = \begin{bmatrix} 0\\ 0\\ 0.5 \end{bmatrix} \end{aligned}\]
  • 约减形式一

如果针对 $A(2,1)$ 进行规则约减，即 $A(2,1)=\sum_ih_iA_i(2,1)\equiv\frac{1+0}{2}=0.5$。此时相当于 $h_{211},h_{212}$ 失去作用，$A_1,A_2$ 合并为新的 $A_1^\prime$（原文中仍以 $A_1$ 表示），其它矩阵类似。隶属度函数 $h_1,h_2$ 合并为 $h_1^\prime$（原文中以 $m_1$ 表示），其它隶属度函数类似。

则规则数和矩阵数目减半，系统变为

\[\begin{aligned} \dot \boldsymbol x(t) &= \sum_{i=1}^4 h_i^\prime(\boldsymbol z(t))[\boldsymbol A_i^\prime\boldsymbol x(t) + \boldsymbol B_i^\prime\boldsymbol u(t)]\\ h_1^\prime &= h_{331}v_{311}=\frac{5+x_1(t)}{10}\cdot\frac{1+{\rm sin}x_3(t)}{2}\\ h_2^\prime &= h_{332}v_{311}=\frac{5-x_1(t)}{10}\cdot\frac{1+{\rm sin}x_3(t)}{2}\\ h_3^\prime &= h_{331}v_{312}=\frac{5+x_1(t)}{10}\cdot\frac{1-{\rm sin}x_3(t)}{2}\\ h_4^\prime &= h_{332}v_{312}=\frac{5-x_1(t)}{10}\cdot\frac{1-{\rm sin}x_3(t)}{2}\\ \boldsymbol A_1^\prime &= \begin{bmatrix} 0&1&0\\ \boldsymbol 0.5&0&-1\\ 0&0&5 \end{bmatrix},\ \boldsymbol A_2^\prime = \begin{bmatrix} 0&1&0\\ \boldsymbol 0.5&0&-1\\ 0&0&-5 \end{bmatrix}\\ \boldsymbol A_3^\prime &= \begin{bmatrix} 0&1&0\\ \boldsymbol 0.5&0&-1\\ 0&0&5 \end{bmatrix},\ \boldsymbol A_4^\prime = \begin{bmatrix} 0&1&0\\ \boldsymbol 0.5&0&-1\\ 0&0&-5 \end{bmatrix}\\ \boldsymbol B_1^\prime &= \boldsymbol B_2^\prime = \begin{bmatrix} 0\\ 0\\ 1.5 \end{bmatrix},\ \boldsymbol B_3^\prime = \boldsymbol B_4^\prime = \begin{bmatrix} 0\\ 0\\ 0.5 \end{bmatrix} \end{aligned}\]

相应的模型不确定性为

\[\vert\vert\boldsymbol \Delta_{ai}(t) \vert\vert \leq 0.5,\ i=1,2,3,4\]
  • 约减形式二

针对 $A(3,3)$ 进行约减，即 $A(3,3)=\sum_ih_iA_i(3,3)\equiv\frac{5+(-5)}{2}=0$，系统变为

\[\begin{aligned} \dot \boldsymbol x(t) &= \sum_{i=1}^4 h_i^\prime(\boldsymbol z(t))[\boldsymbol A_i^\prime\boldsymbol x(t) + \boldsymbol B_i^\prime\boldsymbol u(t)]\\ h_1^\prime &= h_{211}v_{311}={\rm cos}x_2(t)\cdot\frac{1+{\rm sin}x_3(t)}{2}\\ h_2^\prime &= h_{212}v_{311}=(1-{\rm cos}x_2(t))\cdot\frac{1+{\rm sin}x_3(t)}{2}\\ h_3^\prime &= h_{211}v_{312}={\rm cos}x_2(t)\cdot\frac{1-{\rm sin}x_3(t)}{2}\\ h_4^\prime &= h_{212}v_{312}=(1-{\rm cos}x_2(t))\cdot\frac{1-{\rm sin}x_3(t)}{2}\\ \boldsymbol A_1^\prime &= \begin{bmatrix} 0&1&0\\ 1&0&-1\\ 0&0&\boldsymbol 0 \end{bmatrix},\ \boldsymbol A_2^\prime = \begin{bmatrix} 0&1&0\\ 0&0&-1\\ 0&0&\boldsymbol 0 \end{bmatrix}\\ \boldsymbol A_3^\prime &= \begin{bmatrix} 0&1&0\\ 1&0&-1\\ 0&0&\boldsymbol 0 \end{bmatrix},\ \boldsymbol A_4^\prime = \begin{bmatrix} 0&1&0\\ 0&0&-1\\ 0&0&\boldsymbol 0 \end{bmatrix}\\ \boldsymbol B_1^\prime &= \boldsymbol B_2^\prime = \begin{bmatrix} 0\\ 0\\ 1.5 \end{bmatrix},\ \boldsymbol B_3^\prime = \boldsymbol B_4^\prime = \begin{bmatrix} 0\\ 0\\ 0.5 \end{bmatrix} \end{aligned}\]

相应的模型不确定性为

\[\vert\vert\boldsymbol \Delta_{ai}(t) \vert\vert \leq 5,\ i=1,2,3,4\]
  • 约减形式三

针对 $B(3,1)$ 进行约减，即 $B(3,1)=\sum_ih_iB_i(3,1)\equiv\frac{1.5+0.5}{2}=1$，系统变为

\[\begin{aligned} \dot \boldsymbol x(t) &= \sum_{i=1}^4 h_i^\prime(\boldsymbol z(t))[\boldsymbol A_i^\prime\boldsymbol x(t) + \boldsymbol B_i^\prime\boldsymbol u(t)]\\ h_1^\prime &= h_{211}h_{331}={\rm cos}x_2(t)\cdot\frac{5+x_1(t)}{10}\\ h_2^\prime &= h_{212}h_{331}=(1-{\rm cos}x_2(t))\cdot\frac{5+x_1(t)}{10}\\ h_3^\prime &= h_{211}h_{332}={\rm cos}x_2(t)\cdot\frac{5-x_1(t)}{10}\\ h_4^\prime &= h_{212}h_{332}=(1-{\rm cos}x_2(t))\cdot\frac{5-x_1(t)}{10}\\ \boldsymbol A_1^\prime &= \begin{bmatrix} 0&1&0\\ 1&0&-1\\ 0&0&5 \end{bmatrix},\ \boldsymbol A_2^\prime = \begin{bmatrix} 0&1&0\\ 0&0&-1\\ 0&0&5 \end{bmatrix}\\ \boldsymbol A_3^\prime &= \begin{bmatrix} 0&1&0\\ 1&0&-1\\ 0&0&-5 \end{bmatrix},\ \boldsymbol A_4^\prime = \begin{bmatrix} 0&1&0\\ 0&0&-1\\ 0&0&-5 \end{bmatrix}\\ \boldsymbol B_1^\prime &= \boldsymbol B_2^\prime = \begin{bmatrix} 0\\ 0\\ \boldsymbol 1 \end{bmatrix},\ \boldsymbol B_3^\prime = \boldsymbol B_4^\prime = \begin{bmatrix} 0\\ 0\\ \boldsymbol 1 \end{bmatrix} \end{aligned}\]

相应的模型不确定性为

\[\vert\vert\boldsymbol \Delta_{bi}(t) \vert\vert \leq 0.5,\ i=1,2,3,4\]

3.3. 模糊控制器

3.3.1. 设计

3.3.2. 举例

未完待续！

4. 参考文献

无。

Academic, Paper
deep learning fuzzy
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